When a 4 kg mass is hung vertically on a light spring that obeys Hooke’s law, the spring stretches by 2 cms. The work required to be done by an external agent in the stretching this spring by 5 cms will be (g = 9.8 m/s²)

The correct answer is 2.450 J.

To find the work required to stretch the spring by a given distance, we must first determine the spring constant ($k$) using Hooke's law.

According to Hooke's law, the restoring force exerted by a spring is proportional to its displacement. When a mass $m$ is hung vertically on the spring, it stretches by a distance $x$ under the action of gravity. At equilibrium, the gravitational force acting on the mass is balanced by the spring's restoring force:
$F=kx=mg$

From the given data:
Mass, $m=4\text{kg}$
Acceleration due to gravity, $g=9.8{\text{m/s}}^2$
Initial stretch, $x_1=2\text{cm}=0.02\text{m}$

Substitute these values into Hooke's law to solve for the spring constant ($k$):
$k\cdot 0.02=4\cdot 9.8$
$k\cdot 0.02=39.2$
$k=\frac{39.2}{0.02}=1960\text{N/m}$

Now, we calculate the work done ($W$) by an external agent in stretching this spring by a distance $x_2=5\text{cm}=0.05\text{m}$ from its unstretched position. The work done is stored as the elastic potential energy of the spring:
$W=\frac{1}{2}k{x}_2^2$

Substitute the values of $k$ and $x_2$ into the potential energy formula:
$W=\frac{1}{2}\cdot 1960\cdot {\left(0.05\right)}^2$
$W=980\cdot 0.0025$
$W=2.45\text{J}$

Therefore, the work required to stretch the spring by 5 cm is 2.450 J.